Class 9th Science Chapter-7 In-Text Questions

Chapter – 7 Motion

Page No. 74 – Questions & Answers

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Question 1

An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.

Answer:
Yes, an object can have zero displacement even after moving through a distance.

Explanation:
Displacement depends on the initial and final positions of an object. If the final position is the same as the initial position, displacement becomes zero.

Example:
If a person walks 10 m forward and then 10 m backward, the distance travelled is 20 m, but since the person returns to the starting point, the displacement is zero.

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Question 2

A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

Answer:

• Side of square field = 10 m

• Perimeter of square = 4 × 10 = 40 m

• Time to complete 1 round = 40 s

Total time given = 2 min 20 s = 140 s

Number of rounds completed =
140 ÷ 40 = 3.5 rounds

After 3 full rounds, the farmer comes back to the starting point.
After half round, he reaches the opposite corner of the square.

Magnitude of displacement = diagonal of square

$$\text{Displacement}=\sqrt{{10}^2+{10}^2}=\sqrt{200}=10\sqrt{2}\text{ m}$$

Final Answer:
Magnitude of displacement = $10\sqrt{2}$ m

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Question 3

Which of the following is true for displacement?

(a) It cannot be zero.
(b) Its magnitude is greater than the distance travelled by the object.

Answer:

Neither (a) nor (b) is true.

Explanation:

• Displacement can be zero when the initial and final positions are the same.

• The magnitude of displacement is always less than or equal to the distance travelled, never greater.

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Page No. 76 – Questions & Answers

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Question 1

Distinguish between speed and velocity.

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Question 2

Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?

Answer:
The magnitude of average velocity is equal to average speed when the object moves in a straight line without changing direction.

Explanation:
In this case, distance = displacement, so both average speed and average velocity become equal.

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Question 3

What does the odometer of an automobile measure?

Answer:
An odometer measures the total distance travelled by an automobile.

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Question 4

What does the path of an object look like when it is in uniform motion?

Answer:
When an object is in uniform motion, it moves along a straight-line path.

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Question 5

During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station?
(The signal travels at the speed of light = $3\times {10}^8$ m s⁻¹)

Given:
Speed of signal,

$$v=3\times {10}^8{\text{ m s}}^{-1}$$

Time taken,

$$t=5\text{ minutes}=5\times 60=300\text{ s}$$

Distance = Speed × Time

$$s=v\times t$$
$$s=3\times {10}^8\times 300$$
$$s=9\times {10}^{10}\text{ m}$$

Final Answer:
Distance of the spaceship = $9\times {10}^{10}$ m

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Page No. 77 – Questions & Answers

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Question 1

When will you say a body is in
(i) uniform acceleration?
(ii) non-uniform acceleration?

Answer:

(i) Uniform acceleration:
A body is said to be in uniform acceleration when its velocity changes by equal amounts in equal intervals of time, no matter how small the time intervals are.

Example:
A freely falling body under gravity.

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(ii) Non-uniform acceleration:
A body is said to be in non-uniform acceleration when its velocity changes by unequal amounts in equal intervals of time.

Example:
A car moving in heavy traffic where speed changes irregularly.

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Question 2

A bus decreases its speed from 80 km h⁻¹ to 60 km h⁻¹ in 5 s. Find the acceleration of the bus.

Given:
Initial velocity,

$$u=80{\text{ km h}}^{-1}=\frac{80\times 1000}{3600}=22.22{\text{ m s}}^{-1}$$

Final velocity,

$$v=60{\text{ km h}}^{-1}=\frac{60\times 1000}{3600}=16.67{\text{ m s}}^{-1}$$

Time,

$$t=5\text{ s}$$

Formula:

$$a=\frac{v-u}{t}$$

Calculation:

$$a=\frac{16.67-22.22}{5}=\frac{-5.55}{5}=-1.11{\text{ m s}}^{-2}$$

Answer:
Acceleration of the bus = –1.11 m s⁻²
(Negative sign indicates deceleration)

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Question 3

A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km h⁻¹ in 10 minutes. Find its acceleration.

Given:
Initial velocity,

$$u=0$$

Final velocity,

$$v=40{\text{ km h}}^{-1}=\frac{40\times 1000}{3600}=11.11{\text{ m s}}^{-1}$$

Time,

$$t=10\text{ minutes}=600\text{ s}$$

Formula:

$$a=\frac{v-u}{t}$$

Calculation:

$$a=\frac{11.11-0}{600}=0.0185{\text{ m s}}^{-2}$$

Answer:
Acceleration of the train = 0.0185 m s⁻²

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Page No. 81 – Questions & Answers

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Question 1

What is the nature of the distance–time graphs for uniform and non-uniform motion of an object?

Answer:

• Uniform motion:
  The distance–time graph is a straight line because the object covers equal distances in equal intervals of time.

• Non-uniform motion:
  The distance–time graph is a curved line because the object covers unequal distances in equal intervals of time.

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Question 2

What can you say about the motion of an object whose distance–time graph is a straight line parallel to the time axis?

Answer:
The object is at rest.

Explanation:
A straight line parallel to the time axis means that the distance does not change with time, so the object is not moving.

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Question 3

What can you say about the motion of an object if its speed–time graph is a straight line parallel to the time axis?

Answer:
The object is moving with uniform speed (constant speed).

Explanation:
A straight line parallel to the time axis in a speed–time graph indicates that speed remains constant with time, and acceleration is zero.

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Question 4

What is the quantity which is measured by the area occupied below the velocity–time graph?

Answer:
The area under the velocity–time graph represents the displacement of the object.

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Page No. 82–83 : Questions & Answers

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Question 1

A bus starting from rest moves with a uniform acceleration of 0.1 m s⁻² for 2 minutes. Find
(a) the speed acquired
(b) the distance travelled

Given:

Initial velocity,

$$u=0$$

Acceleration,

$$a=0.1{\text{ m s}}^{-2}$$

Time,

$$t=2\text{ min}=120\text{ s}$$

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(a) Speed acquired

Using equation:

$$v=u+at$$
$$v=0+(0.1\times 120)$$
$$v=12{\text{ m s}}^{-1}$$

Speed acquired = 12 m s⁻¹

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(b) Distance travelled

Using equation:

$$s=ut+\frac{1}{2}a{t}^2$$

$$s=0+\frac{1}{2}\times 0.1\times (120{)}^2$$
$$s=0.05\times 14400$$
$$s=720\text{ m}$$

Distance travelled = 720 m

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Question 2

A train is travelling at a speed of 90 km h⁻¹. Brakes are applied to produce a uniform acceleration of –0.5 m s⁻². Find how far the train will go before it is brought to rest.

Given:

Initial velocity,

$$u=90{\text{ km h}}^{-1}=25{\text{ m s}}^{-1}$$

Final velocity,

$$v=0$$

Acceleration,

$$a=-0.5{\text{ m s}}^{-2}$$

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Using equation:

$$v^2=u^2+2as$$
$$0=(25{)}^2+2(-0.5)s$$
$$0=625-s$$
$$s=625\text{ m}$$

Distance travelled before stopping = 625 m

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Question 3

A trolley, while going down an inclined plane, has an acceleration of 2 cm s⁻². What will be its velocity 3 s after the start?

Given:

Initial velocity,

$$u=0$$

Acceleration,

$$a=2{\text{ cm s}}^{-2}$$

Time,

$$t=3\text{ s}$$

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Using equation:

$$v=u+at$$
$$v=0+(2\times 3)$$
$$v=6{\text{ cm s}}^{-1}$$

Velocity after 3 s = 6 cm s⁻¹

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Question 4

A racing car has a uniform acceleration of 4 m s⁻². What distance will it cover in 10 s after start?

Given:

$$u=0,a=4{\text{ m s}}^{-2},t=10\text{ s}$$

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Using equation:

$$s=ut+\frac{1}{2}a{t}^2$$
$$s=0+\frac{1}{2}\times 4\times (10{)}^2$$
$$s=2\times 100$$
$$s=200\text{ m}$$

Distance covered = 200 m

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Question 5

A stone is thrown vertically upwards with a velocity of 5 m s⁻¹. If acceleration is 10 m s⁻² downward, find:
(a) maximum height attained
(b) time taken to reach maximum height

Given:

$$u=5{\text{ m s}}^{-1},v=0,a=-10{\text{ m s}}^{-2}$$

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(a) Maximum height

Using equation:

$$v^2=u^2+2as$$
$$0=25-20s$$
$$s=1.25\text{ m}$$

Maximum height = 1.25 m

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(b) Time taken

Using equation:

$$v=u+at$$
$$0=5-10t$$
$$t=0.5\text{ s}$$

Time taken = 0.5 s